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Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
8
Rotational Motion
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8-1: Angular Quantities
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8-2 and 8-3: Constant Angular Acceleration; Rolling
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8-4: Torque
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8-5 and 8-6: Rotational Dynamics
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8-7: Rotational Kinetic Energy
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8-8: Angular Momentum
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8-9: Angular Quantities as Vectors
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Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 30
Q

Determine the moment of inertia of a 10.8-kg sphere of radius 0.648 m when the axis of rotation is through its center.

A
1.81 kg m21.81 \textrm{ kg m}^2
Giancoli 7th Edition, Chapter 8, Problem 30 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The moment of inertia for a sphere rotating about an axis through its center is two-fifths times its mass times its radius squared. So that's two-fifth times 10.8 kilograms times 0.648 meters and square that meters and you get 1.81 kilogram meters squared for the moment of inertia.

COMMENTS
By taylornguyen75 on Thu, 1/16/2020 - 2:56 PM

how did you get 2/5?
how do I determine the coefficient?

By Mr. Dychko on Fri, 1/17/2020 - 8:24 PM

Hi taylornguyen75, thanks for the question. This formula comes from the table listing moments of inertia for different shapes. The 2/5 is part of the formula for a sphere.
All the best,
Mr. Dychko

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